Problem: There is a light source on the plane. This source is so small that it can be represented as point. The light source is moving from point (a, s_y) to the (b, s_y) (s_y < 0) with speed equal to 1 unit per second. The trajectory of this light source is a straight segment connecting these two points. 

There is also a fence on OX axis represented as n segments (l_i, r_i) (so the actual coordinates of endpoints of each segment are (l_i, 0) and (r_i, 0)). The point (x, y) is in the shade if segment connecting (x,y) and the current position of the light source intersects or touches with any segment of the fence.

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You are given q points. For each point calculate total time of this point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y).

Input

First line contains three space separated integers s_y, a and b (-10^9 ≤ s_y < 0, 1 ≤ a < b ≤ 10^9) — corresponding coordinates of the light source.

Second line contains single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — number of segments in the fence.

Next n lines contain two integers per line: l_i and r_i (1 ≤ l_i < r_i ≤ 10^9, r_{i - 1} < l_i) — segments in the fence in increasing order. Segments don't intersect or touch each other.

Next line contains single integer q (1 ≤ q ≤ 2 ⋅ 10^5) — number of points to check.

Next q lines contain two integers per line: x_i and y_i (1 ≤ x_i, y_i ≤ 10^9) — points to process.

Output

Print q lines. The i-th line should contain one real number — total time of the i-th point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y). The answer is considered as correct if its absolute of relative error doesn't exceed 10^{-6}.

Example

Input

-3 1 6
2
2 4
6 7
5
3 1
1 3
6 1
6 4
7 6


Output

5.000000000000000
3.000000000000000
0.000000000000000
1.500000000000000
2.000000000000000

Note

  * The 1-st point is always in the shade; 
  * the 2-nd point is in the shade while light source is moving from (3, -3) to (6, -3); 
  * the 3-rd point is in the shade while light source is at point (6, -3). 
  * the 4-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and at point (6, -3); 
  * the 5-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and from (5.5, -3) to (6, -3); 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7, abc = 864197532, N = 800007, logN = 17;
const double eps = 1e-9;
vector<double> v;
struct Seg {
  int l, r, m;
  double val = 0;
  Seg* ch[2];
  Seg(int _l, int _r) : l(_l), r(_r), m(l + r >> 1) {
    if (r - l > 1) {
      ch[0] = new Seg(l, m);
      ch[1] = new Seg(m, r);
    }
  }
  void pull() { val = ch[0]->val + ch[1]->val; }
  void add(int p) {
    if (r - l == 1)
      val += v[p + 1] - v[p];
    else
      ch[p >= m]->add(p), pull();
  }
  double query(int a, int b) {
    if (a <= l && r <= b) return val;
    double ans = 0;
    if (a < m) ans += ch[0]->query(a, b);
    if (m < b) ans += ch[1]->query(a, b);
    return ans;
  }
};
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int sy, a, b;
  cin >> sy >> a >> b;
  int n, q;
  cin >> n;
  vector<pair<int, int>> seg(n);
  for (int i = 0; i < n; ++i)
    cin >> seg[i].first >> seg[i].second, v.push_back(seg[i].first),
        v.push_back(seg[i].second);
  cin >> q;
  vector<pair<int, int>> query(q);
  vector<pair<double, double>> Q(q);
  for (int i = 0; i < q; ++i) {
    cin >> query[i].first >> query[i].second;
    double x1 = 1.0 * query[i].first - 1.0 * (query[i].first - a) *
                                           query[i].second /
                                           (query[i].second - sy);
    double x2 = 1.0 * query[i].first - 1.0 * (query[i].first - b) *
                                           query[i].second /
                                           (query[i].second - sy);
    Q[i] = make_pair(x1, x2);
    v.push_back(x1);
    v.push_back(x2);
  }
  sort(v.begin(), v.end());
  v.resize(unique(v.begin(), v.end()) - v.begin());
  Seg root(0, v.size());
  bool is = false;
  auto same = [&](double a, double b) { return abs(a - b) <= eps; };
  for (int i = 0, j = -1; i < v.size(); ++i) {
    if (j + 1 < n && same(seg[j + 1].first, v[i])) j++, is = true;
    if (same(seg[j].second, v[i])) is = false;
    if (is) root.add(i);
  }
  for (int i = 0; i < q; ++i) {
    double x1, x2;
    tie(x1, x2) = Q[i];
    int f = lower_bound(v.begin(), v.end(), x1) - v.begin(),
        t = lower_bound(v.begin(), v.end(), x2) - v.begin();
    cout << fixed << setprecision(10)
         << root.query(f, t) / query[i].second * (query[i].second - sy) << '\n';
  }
}